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Q.

Figure shows a block P of mass m resting on a horizontal smooth floor at a distance l from a rigid wall. Block is pushed toward right by a distance 3l/2  and released. When block passes from its mean position another block of mass m1 is placed on it which sticks to it due to friction. Find the value of m1 so that the combined block just avoids collision with the left wall.

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a

5m4

b

m2

c

2m

d

m4

answer is B.

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Detailed Solution

So velocity of block when passing from its mean position is given as  v=Aω=3l2km[As  ω=km]
If mass m1  is added to it and just after if velocity of combined block becomes, from momentum conservation we have mv=(m+m1)v1
or  v1=m(m+m1)(3l2km)
If this is the velocity of combined block at mean position, it must be given as
v1=Aω1[Now   ω1=km1+m2]
Where A1  and  ω1 are the new amplitude and angular frequency of SHM of the block. It is given that combined block just reaches the left wall thus the new amplitude of oscillation must be l so we have
m(m1+m2).3l2km=l1km+m1 or3m2m+m1=1 or  9m=4m+4m1 or  m1=54m

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